MHB Show Sentences Proof: "Hey Again! (Wasntme)

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The discussion revolves around proving properties of unions in set theory. It establishes that if an element t belongs to a set A, then t is a subset of the union of A. The proof also confirms that the union of an empty set is indeed empty, and the union of a singleton set equals the element itself. A minor critique is offered regarding the use of variable names in the proof, suggesting that a different variable should be used to avoid confusion. Overall, the proof is well-received with constructive feedback for clarity.
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Hey again! (Wasntme)

I want to show the following:

  • If $t \in A$, then $t \subseteq \cup A$
  • $\cup \varnothing=\varnothing$
  • $ \cup \{ a \}=a$
  • Let $x \in t$.

    So, $\exists t (t \in A \wedge x \in t) \rightarrow x \in \cup A$.
    Therefore, $t \subseteq A$.
  • $$x \in \cup \varnothing \leftrightarrow \exists b (b \in \varnothing \wedge x \in b)$$

    The empty set $\varnothing$ contains no elements, so there is no $b$ such that $b \in \varnothing$, so there is no $x$, such that $x \in \cup \varnothing$.

    Therefore, $\cup \varnothing=\varnothing$.
  • $$x \in \cup \{ a \} \leftrightarrow \exists b (b \in \{ a \} \wedge x \in b) \leftrightarrow (b=a \wedge x \in b) \leftrightarrow x \in a$$

    Therefore, $\cup \{ a \}=a$.
Could you tell me if it is right or if I have done something wrong? (Thinking)
 
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Very well! One minor remark about the first problem. You have a set called $t$. In the scope of this proof, it may be considered a constant. I would not call a variable bound by $\exists$ by the same name $t$. The object immediately following $\exists$ is a variable, not a constant, so when you say $\exists t$ you introduce a new variable that happens to have the same name as the existing object $t$. Then, for example, saying "There exists some $t$ (in fact, it is equal to the set $t$ mentioned above)" would be awkward. So I would say
\[
x\in t\land t\in A\to\exists u\;(x\in u\land u\in A)\to x\in\bigcup A.
\]
Saying that the last implication holds by definition of $\bigcup A$ would not hurt, either. But overall, good job.
 
Evgeny.Makarov said:
Very well! One minor remark about the first problem. You have a set called $t$. In the scope of this proof, it may be considered a constant. I would not call a variable bound by $\exists$ by the same name $t$. The object immediately following $\exists$ is a variable, not a constant, so when you say $\exists t$ you introduce a new variable that happens to have the same name as the existing object $t$. Then, for example, saying "There exists some $t$ (in fact, it is equal to the set $t$ mentioned above)" would be awkward. So I would say
\[
x\in t\land t\in A\to\exists u\;(x\in u\land u\in A)\to x\in\bigcup A.
\]
Saying that the last implication holds by definition of $\bigcup A$ would not hurt, either. But overall, good job.

I understand! Thank you very much! (Nod)
 
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